Tuesday, July 17, 2007

const overload functions taking char pointers

Always have two overloaded versions of functions that take
char * and const char * parameters. Declare (but don't define if not needed)
a function that takes const char* as a parameter when you have defined
a function that accepts a non-const char* as a parameter.

#include <iostream>
#include <iomanip>

static void foo (char *s) {
std::cout << "non-const " << std::hex << static_cast <void *>(s) << std::endl;
}
static void foo (char const *s) {
std::cout << "const " << std::hex << static_cast <void const *>(s) << std::endl;
}

int main (void)
{
char * c1 = "Literal String 1";
char const * c2 = "Literal String 1";
foo (c1);
foo (c2);
foo ("Literal String 1");
//*c1 = 'l'; // This will cause a seg-fault on Linux.
std::cout << c2 << std::endl;

return 0;
}

Because of default conversion rule from string literal to char *,
the call to foo using in-place literal goes completely undetected
through the eyes of compiler's type system.

Interestingly enough, the addresses of all the identical string literals
is the same, irrespective of whether it is assigned to const or non-const.
Internally though, they are stored on the const DATA page and modifying
them causes a seg-fault.

8 comments:

Steven said...

Hi there,


Could you explain why the *c1 = 'l' segfaults under Linux? To me it looks like modifying a proper (non const) string?


Thanks!

Sumant said...

"Literal String 1" is the same for all the char pointers, const or non-const. The compiler detects uniqueness of the initializer strings and creates exactly one copy of it in the DATA pages of the process, which are read only. So const and non-const both point to the same const readonly string and therefore an attempt to modify it seg-faults. I think irrespective to what a string literal is assigned, a string literal will always be in the DATA page readonly.

Suman said...

What you ascribe to the uniqueness of the literals is only a part answer.It would crash even if there was only one non const char pointer to a literal and you attempted to modify the literal's value. The Standard defines it to be an Undefined behaviour and hence *anything* might happen. Some compilers have the option of making literals writable (Gcc is a case in point).

The rational behind such a decision can be attributed to a number of reasons. One example is, as you've said, the literals being in the DATA segment -- modifying the same is tantamount to creating a self-modifying program.

Maxim Yegorushkin said...

Sorry, I did not understand how the fact that a string literal converts to char* infers that one needs two overloads for char* and char const*?

Sumant said...

@maxim
When you declare a function taking a non-const char pointer, you are also declaring your intention to potentially change the input parameter. If you really are sure that you would never change the input parameter string use const char pointer instead. Now assuming that you do intend to change the input parameter string, you declared it as a non-const char pointer and someone out there passed in a string literal to it. That is a problem. Your function will try to change the string literal which is undefined by standard. To avoid this from happening you can take some help from the compiler. If you just declare a function of the same name and rest of the signature, taking const char pointer as input parameter, compiler will pick that up when someone out there uses a string literal to call it. const char pointer is a better match than non-const char pointer for a string literal. But because it is undefined (remember you just declared it) program won't link.

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Amber said...

Of course with modern compilers you **will** get a warning:

test.cpp|16 col 17| warning: deprecated conversion from string constant to ‘char*’ [-Wwrite-strings]

so this advice is a bit obsolete

Sumant Tambe said...

Some people still need old compilers. I do.